Let $g$ be a vector-valued function defined by $g(t)=(-2\sin(t+1),5t^2-2t)$. Find $g'(t)$. Choose 1 answer: Choose 1 answer: (Choice A) A $(-2\cos(t+1),10t-2)$ (Choice B) B $\left(2t\cos(t+1),3t\right)$ (Choice C) C $\left(2(t+1)\cdot\sin(t+1),10t^2-2\right)$ (Choice D) D $2t\cos(t+1)+10t-2$
Explanation: $g$ is a vector-valued function. This means it takes one number as an input $(t)$, but it outputs two numbers as a two-dimensional vector. Finding the derivative of a vector-valued function is pretty straightforward. Suppose a vector-valued function is defined as $u(t)=(v(t),w(t))$, then its derivative is the vector-valued function $u'(t)=(v'(t),w'(t))$. In other words, the derivative is found by differentiating each of the expressions in the function's output vector. Recall that $g(t)=(-2\sin(t+1),5t^2-2t)$. Let's differentiate the first expression: $\dfrac{d}{dt}[-2\sin(t+1)]=-2\cos(t+1)$ Let's differentiate the second expression: $\begin{aligned}\dfrac{d}{dt}(5t^2-2t)&=2\cdot5t-2 \\\\&=10t-2\end{aligned}$ Now let's put everything together: $\begin{aligned} g'(t)&=\left(\dfrac{d}{dt}[-2\sin(t+1)],\dfrac{d}{dt}(5t^2-2t)\right) \\\\ &=(-2\cos(t+1),10t-2) \end{aligned}$ In conclusion, $g'(t)=(-2\cos(t+1),10t-2)$.